How would I find f prime a for the function f(x)=4x^-2? I have no clue how to set it up.

Question

amistre64 · Answer

still going to be the power rule .... define the power rule for me.

amistre64 · Answer

notifs didnt tell me you replied to the other one

amistre64 · Answer

what have you learned about finding a derivative so far?  limit stuff, or are you on to the rules

anonymous · Answer

I know how to find the derivative, and I THINK i should be using (f(a+h)-f(a))/h but I don't know how to fit the problem in to it. How do I get rid of the ^-2? would it become ^1/4?

amistre64 · Answer

recall that a^(-n) = 1/(a^n)  that will help us with it

amistre64 · Answer

$$\frac{(x+h)^{-2}-(x)^{-2}}{h}$$
$$\frac{\dfrac{1}{(x+h)^{2}}-\dfrac{1}{x^{2}}}{h}$$
then it works our by adding the fractions and working a conjugate most likely

anonymous · Answer

wouldn't it be 4(x+h)^-2?

amistre64 · Answer

the 4 is a constant, so its really immaterial.  if you want to include it feel free

anonymous · Answer

Ok. Thank you so much!

anonymous · Answer

I think I can work it out on my own for the mos part now.

amistre64 · Answer

[4 f(x)]'  =  4  f'(x)

constants pull out

amistre64 · Answer

good luck

anonymous · Answer

YES I got it! Thank you!